Directions (95-98): Read the following information carefully and answer the given questions. A shopkeeper sold five articles: P, Q, R, S and T. Further partial information in given in table below. | Article | Cost price | % Profit | Marked price | |---|---|---|---| | P | - | 40% | $(10.8b + 3m)$ | | Q | - | - | $(12a - n)$ | | R | - | 20% | - | | S | $(5b - m - n - 48)$ | - | - | | T | $(4a + \frac{m}{2})$ | - | $(13a - \frac{n}{2} + 6)$ | Note: (i) Value of '$n$' is 4 times of larger root of given equation. $K^2 - 32K + 252 = 0$ (ii) Article T is marked up ₹1280 more than its cost price and the difference between marked price and cost price of article T is 28% more than profit earned on article S. (iii) Discount given on article Q is 16.66%, and selling price of S is ₹$(5m + 7n + 4a + 296)$. (iv) Value of '$m$' is twice of missing value in the given sequence. 140, 136, 161, (?), 329, -200 If cost price of P is 87.5% more than that of T, and P is sold after discount of $(2L - 210)\%$, then which of the following is/are possible value of L? I. $1.5 m$ II. $(b - n - 8)$ III. $2.4(b - a)$ IV. 240

### Concept & Equation Solving Strategy This is a comprehensive Data Interpretation puzzle integrating quadratic equations, number series, and profit & loss concepts. We must decode the underlying variables ($n, m, a, b$) first. Key formulas: $$Profit = SP - CP$$ $$Discount\ \% = \frac{MP - SP}{MP} \times 100$$ ### Step-by-Step Solution **1. Find $n$ (from quadratic equation):** * Equation: $K^2 - 32K + 252 = 0$ * Roots are $14$ and $18$ (since $14 \times 18 = 252$ and $14 + 18 = 32$). * The larger root is $18$. Therefore, $n = 4 \times 18 = 72$. **2. Find $m$ (from sequence):** * Sequence: $140, 136, 161, x, 329, -200$ * Differences: $136 - 140 = -4$ ($-2^2$) * $161 - 136 = 25$ ($+5^2$) * Notice the pattern of alternating signs and squares of alternate prime numbers ($2, 5, 11, 17, 23$). * Next difference should be $-11^2 = -121$. So, $x = 161 - 121 = 40$. * Verify with next: $40 + 17^2 = 40 + 289 = 329$. (Matches perfectly) * Therefore, $m = 2 \times 40 = 80$. **3. Find $a$ (from Article T):** * $CP_T = 4a + \frac{80}{2} = 4a + 40$ * $MP_T = 13a - \frac{72}{2} + 6 = 13a - 30$ * Markup is $1280 \implies MP_T - CP_T = 1280$ * $(13a - 30) - (4a + 40) = 1280 \implies 9a - 70 = 1280 \implies 9a = 1350 \implies a = 150$. **4. Find $b$ (from Article S):** * Markup of T ($1280$) is $28\%$ more than Profit on S. * $Profit_S = \frac{1280}{1.28} = 1000$. * $SP_S = 5(80) + 7(72) + 4(150) + 296 = 400 + 504 + 600 + 296 = 1800$. * $CP_S = SP_S - Profit_S = 1800 - 1000 = 800$. * From table, $CP_S = 5b - m - n - 48 = 5b - 80 - 72 - 48 = 5b - 200$. * Equating: $5b - 200 = 800 \implies 5b = 1000 \implies b = 200$. **5. Evaluate Question for L:** * $CP_T = 4(150) + 40 = 640$. * $CP_P$ is $87.5\%$ ($\frac{7}{8}$) more than $CP_T \implies CP_P = 640 \times \frac{15}{8} = 1200$. * $Profit_P = 40\% \implies SP_P = 1200 \times 1.4 = 1680$. * $MP_P = 10.8(200) + 3(80) = 2160 + 240 = 2400$. * Discount given = $2400 - 1680 = 720$. Discount $\%$ = $(\frac{720}{2400}) \times 100 = 30\%$. * Given Discount $\%$ = $2L - 210 \implies 2L - 210 = 30 \implies 2L = 240 \implies L = 120$. **6. Verify Options:** * I. $1.5m = 1.5(80) = 120$ (True) * II. $b - n - 8 = 200 - 72 - 8 = 120$ (True) * III. $2.4(b - a) = 2.4(200 - 150) = 2.4(50) = 120$ (True) * IV. $240 \neq 120$ (False) ### Exam Strategy & Shortcut By finding the base values ($a=150, b=200, m=80, n=72$) accurately on your scratchpad first, both associated questions collapse into standard arithmetic. Memorizing fractional equivalents (like $87.5\% = \frac{7}{8}$) saves massive calculation time on step 5. ### Common Pitfall A frequent error is misinterpreting the number series. Noticing that differences $(-4, 25)$ are squares of prime numbers with alternating signs is non-intuitive. Always test alternating logic if simple differences yield no standard geometric/arithmetic progression. ### Final Answer Therefore, the correct answer is **I, II, and III only**.